Editing Parseval's theorem (section)

== Notation used in engineering ==

In [[electrical engineering]], Parseval's theorem is often written as:

:<math>\int_{-\infty}^\infty \bigl| x(t) \bigr|^2 \, \mathrm{d}t  =  \frac{1}{2\pi} \int_{-\infty}^\infty \bigl| X(\omega) \bigr|^2 \, \mathrm{d}\omega  =  \int_{-\infty}^\infty \bigl| X(2\pi f) \bigr|^2 \, \mathrm{d}f</math>

where <math>X(\omega) = \mathcal{F}_\omega\{ x(t) \}</math> represents the [[continuous Fourier transform]] (in non-unitary form) of <math>x(t)</math>, and <math>\omega = 2\pi f</math> is frequency in radians per second.

The interpretation of this form of the theorem is that the total [[Energy (signal processing)|energy]] of a signal can be calculated by summing power-per-sample across time or spectral power across frequency.  

For [[discrete time]] [[signal (information theory)|signals]], the theorem becomes:

:<math>\sum_{n=-\infty}^\infty \bigl| x[n] \bigr|^2 = \frac{1}{2\pi} \int_{-\pi}^\pi \bigl| X_{2\pi}({\phi}) \bigr|^2 \mathrm{d}\phi</math>

where <math>X_{2\pi}</math> is the [[discrete-time Fourier transform]] (DTFT) of <math>x</math> and <math>\phi</math> represents the [[angular frequency]] (in [[radian]]s per sample) of <math>x</math>.

Alternatively, for the [[discrete Fourier transform]] (DFT), the relation becomes:

:<math> \sum_{n=0}^{N-1} \bigl| x[n] \bigr|^2  = \frac{1}{N} \sum_{k=0}^{N-1} \bigl| X[k] \bigr|^2</math>

where <math>X[k]</math> is the DFT of <math>x[n]</math>, both of length <math>N</math>.

We show the DFT case below. For the other cases, the proof is similar. By using the definition of inverse DFT of <math>X[k]</math>, we can derive

:<math>\begin{align}
\frac{1}{N} \sum_{k=0}^{N-1} \bigl| X[k] \bigr|^2
&= \frac{1}{N} \sum_{k=0}^{N-1} X[k]\cdot X^*[k]
 = \frac{1}{N} \sum_{k=0}^{N-1} \Biggl(\sum_{n=0}^{N-1} x[n]\,\exp\left(-j\frac{2\pi}{N}k\,n\right)\Biggr) \, X^*[k]
\\[5mu]
&= \frac{1}{N} \sum_{n=0}^{N-1} x[n] \Biggl(\sum_{k=0}^{N-1} X^*[k]\,\exp\left(-j\frac{2\pi}{N}k\,n\right)\Biggr) 
 = \frac{1}{N} \sum_{n=0}^{N-1} x[n] \bigl(N \cdot x^*[n]\bigr)
\\[5mu]
&= \sum_{n=0}^{N-1} \bigl| x[n] \bigr|^2,
\end{align}</math>

where <math>*</math> represents complex conjugate.